Lab Assignment 9

Collaboration Policy

The pre-lab requirement must be completed and submitted individually.

The remainder of the lab activity should be completed working in pairs. One person should submit the result, making sure that both partners' names are clearly identified in that submission.

Please make sure you adhere to the policies on academic integrity in this regard.

Pre-Lab Requirement

Read the complete problem description and then determine what the expected output should be if given the following input:

Prelab input:	Prelab output:
dddududuuu dduudduduu duddduduuu #

Tree Grafting 2

Trees have many applications in computer science. Perhaps the most commonly used trees are rooted binary trees, but there are other types of rooted trees that may be useful as well. One example is ordered trees, in which the subtrees for any given node are ordered. The number of children of each node is variable, and there is no limit on the number. Formally, an ordered tree consists of a finite set of nodes T such that

• there is one node designated as the root, denoted root(T);
  
• the remaining nodes are partitioned into subsets T1, T2, ..., Tm, each of which is also a tree (subtrees).

It is often more convenient to represent an ordered tree as a rooted binary tree, so that each node can be stored in the same amount of memory. The conversion is performed by the following steps:

1. remove all edges from each node to its children;
2. for each node, add an edge to its first child in T (if any) as the left child;
3. for each node, add an edge to its next sibling in T (if any) as the right child.

This is illustrated by the following:

         0                             0
       / | \                          /
      1  2  3       ===>             1
        / \                           \
       4   5                           2
                                      / \
                                     4   3
                                      \
                                       5

You are asked to write a program that computes the height of the tree before and after the conversion.

Input

The input is given by a number of lines giving the directions taken in a depth-first traversal of the trees. There is one line for each tree. For example, the tree above would give dudduduudu, meaning 0 down to 1, 1 up to 0, 0 down to 2, etc. The input is terminated by a line whose first character is #. You may assume that each tree has at least 2 and no more than 10000 nodes.

Output

For each tree, print the heights of the tree before and after the conversion specified above. Use the format:

Tree t: h1 => h2

Hints

Upon first glance, this problem seems like it's going to require you to build a tree data structure and then implement the conversion from the general tree to the binary tree. But as is the case with many of these contest problems, there's a much easier way.

It is possible to compute the resulting depths without ever building the trees. We suggest a recursive approach. In the end we want a function that can compute the depth of the entire tree; we will design a recursion that computes the depth of any subtree. Consider the following local view:

If you were told the traditional depths for each of the four children, can you compute the traditional depth of the larger subtree? Consider the "converted" case. If you were told the "converted" depth for each of the four children, can you determine the converted depth for the larger subtree?

Here is how we recommend setting up the recursion. Remember that the original input is just a sequence of 'd' and 'u' designators to describe the shape of the tree. We want to think about starting the recursion at the root. Technically, there is no 'd'/'u' pair in the input for going down to the root up up from it, but for consistency sake, we recommend that you alter the original input string by adding a preceding 'd' and a trailing 'u'. Then use a string iterator to implicitly traverse the tree as you do your computation. Use the following signature.

int depth(string::iterator& i, bool converted = false);

Finally, we've suggested a signature that computes the traditional depth by default, but can be changed to compute the "converted" depth by sending a second parameter. Although we could have given two different functions for the two different depths, we are suggesting a single function because there is really only a one or two line difference between the two computations. Still, we recommend that you start by implementing the traditional depth version. Then you can go back and do the converted depth. Note that you will make two separate calls from the main routine, and you will need to reinitialize the iterator to the beginning of the string for the second computation.